Question

Find the equation of the plane containing the line \[\frac{x + 1}{- 3} = \frac{y - 3}{2} = \frac{z + 2}{1}\]  and the point (0, 7, −7) and show that the line  \[\frac{x}{1} = \frac{y - 7}{- 3} = \frac{z + 7}{2}\] also lies in the same plane.

 

Answer 1

\[\text{ Let the equation of the plane passing through (0, 7, -7) be } \]
\[a \left( x - 0 \right) + b \left( y - 7 \right) + c \left( z + 7 \right) = 0 . . . \left( 1 \right)\]
\[\text{ The line } \frac{x + 1}{- 3}=\frac{y - 3}{2}=\frac{z + 2}{1} \text{ passes through (-1, 3, -2) and its direction ratios are proportional to -3, 2, 1 } .\]
\[ \text{ Since    plane  (1) contains this line, it must pass through the point (-1, 3, -2) }.\]
\[ \Rightarrow a \left( - 1 - 0 \right) + b \left( 3 - 7 \right) + c \left( - 2 + 7 \right) = 0 \]
\[ \Rightarrow - a - 4b + 5c = 0\]
\[ \Rightarrow a + 4b - 5c = 0 . . . \left( 2 \right)\]
\[\text{ Since    plane (1) contains this line, it must be parallel to the line.} \]
\[ \Rightarrow - 3a + 2b + c = 0 . . . \left( 3 \right)\]
\[ \text{ Solving (1), (2) and (3), we get} \]
\[\begin{vmatrix}x - 0 & y - 7 & z + 7 \\ 1 & 4 & - 5 \\ - 3 & 2 & 1\end{vmatrix} = 0\]
\[ \Rightarrow 14x + 14 \left( y - 7 \right) + 14 \left( z + 7 \right) = 0\]
\[ \Rightarrow 14x + 14y + 14z = 0\]
\[ \Rightarrow x + y + z = 0\]